# Question 4f54f

Apr 25, 2017

${K}_{c} = 0.0062$

#### Explanation:

For starters, calculate the number of moles of hydrogen iodide present in the equilibrium mixture by using the compound's molar mass

1.9 color(red)(cancel(color(black)("g"))) * "1 mole HI"/(127.9color(red)(cancel(color(black)("g")))) = "0.01486 moles HI"

Next, calculate the initial concentration and the equilibrium concentration of hydrogen iodide by using the volume of the container

["HI"]_ 0 = "0.0172 moles"/"2 L" = "0.00860 M"

["HI"]_ "equi" = "0.01486 moles"/"2 L" = "0.00743 M"

Now, the equilibrium reaction looks like this

$\textcolor{red}{2} {\text{HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I}}_{2 \left(g\right)}$

You know that you started with $\text{0.00860 M}$ of hydrogen iodide and ended up with $\text{0.00743 M}$. The difference between these two values represents the concentration of hydrogen iodide that has been converted to hydrogen gas and iodine gas.

["HI"] _ "react" = "0.00860 M" - "0.00743 M"

["HI"]_ "react" = "0.00117 M"

Notice that it takes $\textcolor{red}{2}$ moles of hydrogen iodide to produce $1$ mole of hydrogen gas and $1$ mole of iodine gas. This means that the equilibrium concentrations of the two products will be half the concentration of hydrogen iodide that reacted.

["H"_ 2]_ "equi" = ["HI"]_ "react"/color(red)(2) = "0.00117 M"/color(red)(2) = "0.000585 M"

["I"_ 2]_ "equi" = ["HI"]_ "react"/color(red)(2) = "0.00117 M"/color(red)(2) = "0.000585 M"

By definition, the equilibrium constant for this reaction takes the form

K_c = (["H"_ 2]_ "equi" * ["I"_ 2]_ "equi")/(["HI"]_"equi"^color(red)(2))#

Plug in your values to find

${K}_{c} = \left(0.000585 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{M"))) * 0.000585color(red)(cancel(color(black)("M"))))/(0.00743^color(red)(2) color(red)(cancel(color(black)("M}}^{\textcolor{red}{2}}}}}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.0062}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the container.

Finally, does the result make sense?

Notice that most of the hydrogen iodide that was initially placed in the container remains unreacted. This tells you that ${K}_{c} < 1$, i.e. the equilibrium lies to the left at the temperature at which the reaction takes place.